Problem 34
Question
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through (-2,-5) and (6,-5)
Step-by-Step Solution
Verified Answer
The equation of the line in point slope form is y = -5 and in slope intercept form is also y = -5.
1Step 1: Identify the Coordinates of the Points
The exercise gives two points through which the line passes: Point 1 as (-2, -5) and Point 2 as (6, -5).
2Step 2: Calculate the Slope
The slope is calculated as the change in y divided by the change in x. Here both points have the same y-coordinate(-5), so the change in y is 0, making the slope 0.
3Step 3: Write the Point-slope form
With a slope of 0 and using Point 1 coordinates for (x1, y1), the equation in point-slope form becomes (y - (-5)) = 0*(x - (-2)), which simplifies to y = -5.
4Step 4: Write the Slope-intercept form
The slope is 0 and y-intercept is the y-coordinate of either point, so the equation in slope-intercept form is y = 0*x - 5, which simplifies to y = -5.
Key Concepts
Point-Slope FormSlope-Intercept FormSlope Calculation
Point-Slope Form
Linear equations can be represented in several forms, and the point-slope form is one of them. It is especially useful when you know a point on the line and the slope.
The general formula for point-slope form is given as: \[ y - y_1 = m(x - x_1) \] Here, \( (x_1, y_1) \) is a point on the line, and \( m \) is the slope. Using the point-slope form, you can directly write the equation of the line provided you have these two pieces of information.
In our exercise, we identified the point \((-2, -5)\) and calculated the slope to be 0. Using these, we plug into the formula:
The general formula for point-slope form is given as: \[ y - y_1 = m(x - x_1) \] Here, \( (x_1, y_1) \) is a point on the line, and \( m \) is the slope. Using the point-slope form, you can directly write the equation of the line provided you have these two pieces of information.
In our exercise, we identified the point \((-2, -5)\) and calculated the slope to be 0. Using these, we plug into the formula:
- \( y - (-5) = 0(x - (-2)) \) simplifies to \( y + 5 = 0 \)
- Further simplifies to \( y = -5 \)
Slope-Intercept Form
Another convenient way to express a linear equation is in slope-intercept form. This format emphasizes the slope and the point where the line crosses the y-axis, called the y-intercept.
The slope-intercept form is written as: \[ y = mx + b \] Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept.
For our problem, since the slope is 0 and since the line passes through the y-value of -5 (horizontal line), the equation becomes:
The slope-intercept form is written as: \[ y = mx + b \] Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept.
For our problem, since the slope is 0 and since the line passes through the y-value of -5 (horizontal line), the equation becomes:
- \( y = 0 \times x - 5 \)
- Simplifies to \( y = -5 \)
Slope Calculation
Understanding how to calculate the slope is a crucial step in handling linear equations. The slope is a measure of how "steep" the line is and is commonly represented by \( m \).
It involves comparing the change in y-values to the change in x-values between two points on a line. The formula to calculate slope is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]This formula provides a ratio to describe the tilt of the line.
In our specific example, both given points, \((-2, -5)\) and \((6, -5)\), share the same y-value. Hence, \( y_2 - y_1 = 0 \), leading to a slope \( m = 0 \).
It involves comparing the change in y-values to the change in x-values between two points on a line. The formula to calculate slope is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]This formula provides a ratio to describe the tilt of the line.
In our specific example, both given points, \((-2, -5)\) and \((6, -5)\), share the same y-value. Hence, \( y_2 - y_1 = 0 \), leading to a slope \( m = 0 \).
- When the slope is 0, the line is horizontal.
- No matter how \( x \) changes, \( y \) remains constant.
Other exercises in this chapter
Problem 34
Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=x-6, g(x)=5 x^{2}$$
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Evaluate each function at the given values of the independent variable and simplify. \(f(r)=\sqrt{25-r}-6\) a. \(f(16)\) b. \(f(-24)\) c. \(f(25-2 x)\)
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Write the standard form of the equation of the circle with the given center and radius. Center \((-1,4), r=2\)
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Let \(P(x, y)\) be a point on the graph of \(y=x^{2}-4 .\) Express the distance, \(d,\) from \(P\) to the origin as a function of the point's \(x\) -coordinate.
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